Question

Prove that
$\frac{1}{{log}_{ab}\left(abc\right)}+\frac{1}{{log}_{bc}\left(abc\right)}+\frac{1}{{log}_{ca}\left(abc\right)}=2$

Answer 1

$\frac{1}{lo{g}_{ab}\left(abc\right)}+\frac{1}{lo{g}_{bc}\left(abc\right)}+\frac{1}{lo{g}_{ca}\left(abc\right)}$

$\Rightarrow \frac{logab}{logabc}+\frac{logbc}{logabc}+\frac{logca}{logabc}$ $[lo{g}_{b}a=\frac{lo{g}_{c}a}{lo{g}_{c}b}]$

$\Rightarrow \frac{logab+logbc+logca}{logabc}$

$\Rightarrow \frac{loga+logb+logb+logc+logc+loga}{logabc}\dots (\log a+\log b=\log (ab\left)\right)$

$\Rightarrow \frac{log(abc{)}^2}{log\left(abc\right)}$

$\Rightarrow \frac{2logabc}{logabc}\dots (n{\log }_{a}b={\log }_a{b}^n)$

$\Rightarrow 2$

$\therefore \frac{1}{lo{g}_{ab}\left(abc\right)}+\frac{1}{lo{g}_{bc}\left(abc\right)}+\frac{1}{lo{g}_{ca}\left(abc\right)}=2$

Hence, proved.